Answer: We can use the properties of the normal distribution to solve both parts of this problem.
a) The number of months with sales greater than 100 is a binomial random variable with parameters n = the number of months and p = the probability of a single month having sales greater than 100. Since each month is an independent normal random variable, we can use the standard normal distribution to find this probability. Let X be the number of months with sales greater than 100. Then:
X ~ Binomial(n, p)
where n is the number of months and p is given by:
p = P(X_i > 100) = P(Z > (100 - a) / a)
where Z is a standard normal random variable.
Using the binomial probability formula, the probability of exactly k months having sales greater than 100 is:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
So the probability of exactly 3 months having sales greater than 100 is:
P(X = 3) = (n choose 3) * p^3 * (1 - p)^(n - 3)
= (12 choose 3) * P(Z > (100 - a) / a)^3 * [1 - P(Z > (100 - a) / a)]^(12 - 3)
b) The total sales in the next 12 months is a normal random variable with mean 12a and standard deviation sqrt(12)a. Let Y be the total sales in the next 12 months. Then:
Y ~ Normal(12a, sqrt(12)a)
The probability of the total sales being greater than some value x can be found by standardizing Y and using the standard normal distribution:
P(Y > x) = P((Y - 12a) / (sqrt(12)a) > (x - 12a) / (sqrt(12)a))
= P(Z > (x - 12a) / (sqrt(12)a))
where Z is a standard normal random variable. So the probability of the total sales in the next 12 months being greater than 1500 is:
P(Y > 1500) = P(Z > (1500 - 12a) / (sqrt(12)a))
Explanation: